Homomorphism ABSTRACT ALGEBRA!!!?

Question

Given that m divides n, how do I find a homomorphism where phi(Z_n) = Z_m ? Please explain...

Please if you have no idea what a homomorphism is, keep your stupid comments to yourself!!!

I did not understand your answer, hence why I reposted it....

Answer 1

See my answer to the exact same question which was asked yesterday:

http://answers.yahoo.com/question/index;_ylt=AoeADP1supNGgEBtqBj2LMHsy6IX;_ylv=3?qid=20071215154519AA4LY9Q

Edit: It is difficult to clarify unless you tell me what part of the answer you didn't understand. If it's the coset notation, that comes from the fact that the canonical definition of Z_n is Z/nZ, which is the set of equivalence classes of integers under the relation ~, defined by a~b ⇔ n | a-b. Note that each such equivalence class will have the form a + nZ = {a+nz: z∈Z} for some integer a.

The homomorphism constructed basically consists of associating each integer with the equivalence class of its remainder mod m, with some exposition to handle the fact that each element of Z/nZ is an equivalence class of integers rather than an integer itself, with some exposition to handle the fact that the elements of Z/nZ are equivalence classes of integers rather than integers, so some exposition was required to show that that definition was independent of the representative of each equivalence class chosen.

Actually, in retrospect, I realize that the proof could have been simplified somewhat. Since I _am_ working with equivalence classes rather than individual integers, I could have just assigned the equivalence class of a in Z_n to the equivalence class of a in Z_m with no loss of generality, and that would have significantly simplified the proof. Here's a proof of that approach:

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Define φ: Z_n → Z_m by φ(a+nZ) = a+mZ. We must check that this is well-defined. Suppose a+nZ = b+nZ, then a-b∈nZ, so n | a-b. But then since m | n, we have m | a-b, so a+mZ = b+mZ. Thus φ(a+nZ) is independent of which representative is chosen, so this is well-defined.

That this is a homomorphism is completely obvious: φ((a+nZ) + (b+nZ)) = φ((a+b) + nZ) = (a+b) + mZ = (a+mZ) + (b+mZ) = φ(a+nZ) + φ(a+mZ), so addition is preserved. Similarly, φ((a+nZ) * (b+nZ)) = φ((ab) + nZ) = (ab) + mZ = (a+mZ) * (b+mZ) = φ(a+nZ) * φ(a+mZ), so multiplication is preserved.

Finally, since every element of Z_m has the form a+mZ for some integer a, and a+mZ = φ(a+nZ), it follows that φ is surjective, so we are done.